Let O be the centre of a circle and AB, CD be the two chords. Let PQ be the diameter bisecting chord AB and CD at L and M respectively.
L is the mid- point of AB
So OL ⊥ AB ⇒ ∠ALO = 90 ̊
Similarly, ∠CMO = 90 ̊
∠ALO = ∠CMO = 90 ̊
... co-interior angles are supplementary
So AB ∥ CD.
Hence proved.
Hope this answer helps you.